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A bag contains $4$ red and $3$ black balls.A second bag contains $2$ red and $4$ black balls.If one bag is selected at random and a ball is drawn from it,find probability that the ball is red.

My attempt:- P(Red ball from bag 1)=4/7 P(Red ball from bag 2)=2/6 P(Selecting one bag from 2 bags)=1/2.

So,P(Red ball from bag 1 U Red ball from bag 2) =4/7+2/6=38/42.

So,our required probability from any one bag at random=19/42.

Am I correct?

3 Answers 3

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Using the law of total probability and then Bayes rule, we can write the probability of picking a red ball as

\begin{align*} P(R) &= P(R \cap \text{picked bag 1}) + P(R \cap \text{picked bag 2}) \\ &= P(R \mid \text{picked bag 1})P(\text{picked bag 1}) + P(R \mid \text{picked bag 2})P(\text{picked bag 2}) \\ &= \frac{4}{7} \cdot \frac{1}{2} + \frac{2}{6} \cdot \frac{1}{2} = \frac{2}{7} + \frac{1}{6} = \frac{19}{42} \end{align*}

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    You have got a typo there..it should be 2/6 not2/4 – 2017-01-05
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Assuming that the bags have equal probability of being picked, that is $P(1)=P(2)=\frac{1}{2}$ where $P(i)$ is the probability of picking bag $i$, then you simply need to condition your probabilities

$$P(\text{Red}) = P(\text{Red} | 1) + P(\text{Red}|2)$$

Where $P(A|B)$ is the probability of $A$ given $B$. The above equation is therefore

$$P(\text{Red}) = \frac{1}{2}\frac{4}{7} + \frac{1}{2}\frac{2}{6}$$

I trust you can handle it now :)

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    Ok...thanks...so we have got the same answer...but is my method correct? – 2017-01-05
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    Both methods look the same...adding the probabilities and then multuplying by 1/2 – 2017-01-05
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    To be precise one should state $P(Red)=P(Red\cap 1)+P(Red\cap 2)$ instead of your first equation and then using bayes theorem you derive the second one. – 2017-01-05
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Your answer is correct. But I would be a little more careful about how you get there! We want to condition the event of getting a red ball on which bag it comes from

$$ P(red)=P(red | bag 1)P(bag 1) + P(red | bag 2)P(bag 2) $$

In your particular example $P(bag1)=P(bag2)=\frac 12$ so you can add the probabilities and then multiply the sum by $\frac 12$. If Bag 1 and Bag 2 were chosen with differing probabilities, then your method would break down.